# Disaggregation of a random variable

Assume a random variable $$Z$$ which is the sum of many non-observable random variables, say $$Z = X_1 + \dotsm +X_n$$ for which I know the probability density $$p(X)$$. Now given an observation $$Z=z$$, my problem is to find the most probable vector $$(x_1,\dotsc,x_n)$$ that generates that $$z$$. I would compute the mode of conditional joint probability density $$p(x_1,x_2,\dotsc,x_n \mid Z=z)$$. At the moment I need an expression for this conditional joint expression, all the material I have consulted so far didn't help much.

• Using bayes rule, you can show that it is equal to 0 when $x_1 + x_2 + \ldots + x_n \neq z$ and equal to $\frac{\prod_i p(x_1)}{(\ast^n p)(z)}$ when the equality holds, where $(\ast^n p)$ is the shorthand for n-fold convolution of the marginal of $X$. – Innuo Oct 15 '15 at 14:04
• @Innuo, are you assuming X are independent? How would change the formula if they are not? – Davide C Oct 19 '15 at 12:39

## 1 Answer

Since $z=x_1+\ldots+x_n$, the density of $(X_1,\ldots,X_{n-1},Z)$ is $$p(x_1)\cdots p(x_{n-1})p(z-x_1-\ldots-x_{n-1})$$ which is proportional to the conditional density of $(X_1,\ldots,X_{n-1}$ given $Z=z$, which is also the conditional density of $(X_1,\ldots,X_{n}$ given $Z=z$ since $X_n$ is a deterministic function of the others.